Question

What is the pH of the solution that results from micing 75 mL of 0.50 M NH₃(aq) and 75 mL of 0.50 HCL(aq) at 25 xC? (K_b for NH₃ = 1.8x 10^-5)

a. 0.60

b 2.67

c. 4.74

d. 4.93

e. 9.26

Answer 1

answer : d. 4.93

solution :

millimoles of NH3 = 75 x 0.5 = 37.5

millimoles of HCl = 75 x 0.5 = 37.5

NH3 + HCl --------------------->NH4Cl

37.5      37.5                           0   --------------------> initial

0            0                               37.5 ----------------------> after reaction

here only the salt NH4Cl remained in the solution

NH4Cl concentration = 37.5 / total volume

                                    =37.5 / (75+75)

                                    = 0.25 M

NH4Cl is a salt of strong acid and weak base . so we can use salt hydrolysis pH formula. the pH should less than 7

pH = 7 - 1/2 [pKb -log C]

pH = 7 - 1/2 [4.74 -log (0.25)]

pH = 4.93